It might have been
$${\mathtt{a}} = {\frac{\left(\left({{\mathtt{8}}}^{-{\mathtt{2}}}\right){\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{7}}}\right)}{{{\mathtt{2}}}^{{\mathtt{6}}}}} \Rightarrow {\mathtt{a}} = {\mathtt{0.031\: \!25}}$$
$${\mathtt{4\,012}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{3}} = {\mathtt{4\,018}}$$
$${\mathtt{\,-\,}}{\mathtt{3}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{\,\small\textbf+\,}}{\mathtt{18}} = {\mathtt{\,-\,}}{\mathtt{3}}{n}{\mathtt{\,\small\textbf+\,}}{\mathtt{18}}{\mathtt{\,\small\textbf+\,}}$$
You said ketchup not catch up
$${\mathtt{a}} = {{\mathtt{8}}}^{{\mathtt{\,-\,}}\left({{\mathtt{2.2}}}^{{\mathtt{7}}}\right)} \Rightarrow {\mathtt{a}} = {\mathtt{0}}$$
$${\mathtt{b}} = {{\mathtt{4}}}^{-{\mathtt{3}}} \Rightarrow {\mathtt{b}} = {\mathtt{0.015\: \!625}}$$
Hit log then type 1/2 then (9) then enter
$${\frac{{\mathtt{log1}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}\left({\mathtt{9}}\right) = {\frac{{log}_{10}\left({\mathtt{1}}\right)}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\mathtt{9}}$$
) 